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2014 AMC 10B Problem 6

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2014 AMC 10B Problem 6AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 2 of 5 · about 2 min · position P1-10

Orvin went to the store with just enough money to buy 3030 balloons. When he arrived, he discovered that the store had a special sale on balloons: buy 11 balloon at the regular price and get a second at 13\frac{1}{3} off the regular price. What is the greatest number of balloons Orvin could buy?

  1. A)

    3333

  2. B)

    3434

  3. C)

    3636

  4. D)

    3838

  5. E)

    3939

Answer

C

Key insight

Under the sale a pair of balloons costs 1 + 2/3 = 5/3 of a regular price, so 30 prices buy 18 pairs.

Solution

Let one balloon regularly cost 11 unit, so Orvin has 3030 units.

Under the sale, a pair of balloons costs 1+(1−13)=1+23=531 + \left(1 - \frac13\right) = 1 + \frac23 = \frac53 units. The number of pairs he can afford is
305/3=30⋅35=18,\frac{30}{5/3} = 30 \cdot \frac35 = 18,
which is 2⋅18=362 \cdot 18 = 36 balloons, using all his money exactly.

The answer is (C) 36\boxed{\textbf{(C)}\ 36}.

Why this works

Choose the regular price as the unit; then the budget is simply the number of balloons it buys at full price. A "buy one, get one discounted" sale is best analyzed per pair, since the pair is the repeating unit of the deal. Also note that a leftover single balloon would be at full price, so check whether the budget divides evenly (here it does).

The trap

Treating the second balloon as costing 1/3 of the price instead of 2/3 of it (1/3 off), which gives 45 balloons.

Common mistakes

  • Treating the second balloon as costing 1/3 of the price instead of 2/3 of it (1/3 off), which gives 45 balloons.
  • Averaging carelessly and using 56\frac{5}{6} per balloon on 3030 balloons instead of on 3030 units of money.

Techniques

Set up the equation/formula and compute; no special trick needed